50.05
Molar mass of CaCO3 = 66.1221g CaCO3/mole CaCO3. This means that 1 mole CaCO3 = 66.1221g CaCO3. To find the mass of 4.5 mole CaCO3, complete the following calculation: 4.5g CaCO3 X 1mol CaCO3/66.1221g CaCO3 = 0.068 mole CaCO3.
Both C7H16 and CaCO3 have a molar mass of approximately 116 grams per mole.
250 grams CaCO3 (1 mole CaCO3/100.09 grams) = 2.50 moles of calcium carbonate
To calculate the amount of calcium in 34.5 g of CaCO3, you need to consider the molar mass of CaCO3 which is 100.09 g/mol. Calcium accounts for approximately 40.08 g in every 100.09 g of CaCO3, which means there are (40.08/100.09) * 34.5 g = 13.82 g of calcium in 34.5 g of CaCO3.
To calculate the mass of lime produced, we need to consider the chemical equation for the reaction: CaCO3 (limestone) -> CaO (lime) + CO2 The molar mass of CaCO3 is 100.09 g/mol and that of CaO is 56.08 g/mol. In this reaction, 1 mole of CaCO3 produces 1 mole of CaO. First, convert 1.5 x 10^3 kg of limestone to grams, then divide by the molar mass of CaCO3 to find the number of moles. Finally, multiply the moles of CaCO3 by the molar ratio of CaCO3 to CaO to find the mass of lime produced.
Molar mass of CaCO3 = 66.1221g CaCO3/mole CaCO3. This means that 1 mole CaCO3 = 66.1221g CaCO3. To find the mass of 4.5 mole CaCO3, complete the following calculation: 4.5g CaCO3 X 1mol CaCO3/66.1221g CaCO3 = 0.068 mole CaCO3.
Both C7H16 and CaCO3 have a molar mass of approximately 116 grams per mole.
CaCO3 +2HCl ------------> CaCl2 + CO2 + H2O number of moles of CO2 in .44 grams = .44/ 44 = .01 From equation it is clear that 1 mole of CO2 is produced from CaCO3 = 1 mole .01 mole of CO2 is formed from CaCO3 = .01 mole Weight of .01 mole of CaCO3 is = .01mole *100 g/mole = 1 gram weight % of CaCO3 is = 1*100/ 1.25 = 80 % w/w I've post my answer, so why don't you show that answer here with the question. It's fare. I must be informed about my answer weather it is right or wrong. Please inform me at amitmahalwar@yahoo.com
To convert alkalinity (HCO3) to CaCO3, you need to use the molar mass ratio. For every mole of bicarbonate (HCO3), you have one mole of carbonate (CO3) in CaCO3. So, to convert, you can multiply the HCO3 concentration by a factor of 50.04 (molar mass of CaCO3/molar mass of HCO3).
250 grams CaCO3 (1 mole CaCO3/100.09 grams) = 2.50 moles of calcium carbonate
To calculate the amount of calcium in 34.5 g of CaCO3, you need to consider the molar mass of CaCO3 which is 100.09 g/mol. Calcium accounts for approximately 40.08 g in every 100.09 g of CaCO3, which means there are (40.08/100.09) * 34.5 g = 13.82 g of calcium in 34.5 g of CaCO3.
The balanced chemical equation for the production of carbon dioxide from calcium carbonate is CaCO3 → CaO + CO2. From this equation, it can be seen that one mole of CaCO3 produces one mole of CO2. At STP, one mole of any gas occupies 22.4 L. Therefore, 55.0 L of CO2 corresponds to 55.0/22.4 = 2.46 moles of CO2. Since one mole of CaCO3 produces one mole of CO2, 2.46 moles of CO2 would require 2.46 moles of CaCO3. Finally, the molar mass of CaCO3 is approximately 100 g/mol, so 2.46 moles of CaCO3 would be 2.46 * 100 = 246 grams of CaCO3 are needed.
To calculate the mass of lime produced, we need to consider the chemical equation for the reaction: CaCO3 (limestone) -> CaO (lime) + CO2 The molar mass of CaCO3 is 100.09 g/mol and that of CaO is 56.08 g/mol. In this reaction, 1 mole of CaCO3 produces 1 mole of CaO. First, convert 1.5 x 10^3 kg of limestone to grams, then divide by the molar mass of CaCO3 to find the number of moles. Finally, multiply the moles of CaCO3 by the molar ratio of CaCO3 to CaO to find the mass of lime produced.
CaCO3 + 2HCl -> CaCl2 + CO2 + H2O 1 tonne = 2000 lbs ( I guess this is what you mean ) 2000 lbs CaCO3 (454 grams/1 lb)(1 mole CaCO3/100.09 grams)(1 mole H2O/1 mole CaCO3)(18.016 grams/1 mole H2O) = 163438.1856 grams of water ---------------------------- 163438.1856 grams (1 lb/454 grams) = 360 pounds of water -------------------------------
0.50 moles CaCO3 (1 mole Ca/1 mole CaCO3)(6.022 X 1023/1 mole Ca)= 3.0 X 1023 atoms of calcium===================
1 mole CaCO3 (3 mole O/1 mole CaCO3) = 3 moles oxygen
There are 2 oxygen atoms in one molecule of CaCO3. To calculate the number of oxygen atoms in 50 grams of CaCO3, you first need to find the number of moles of CaCO3 using its molar mass. Then, multiply the number of moles by the number of atoms of oxygen per molecule of CaCO3 (2) to find the total number of oxygen atoms.